WEF Discussions — Now on LinkedIn!

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Yes we use Bio P removal and it works good. For 2009 the average was 8.39 MGD flow, 3.71 mg/L-P influent and 0.34 mg/L-P effluent. The plant design is 8.0 MGD. We have 2 anaerobic tanks and 5 aerobic tanks, 5 clarifiers, nitrification towers for ammonia, and cloth disk filters. At design flow the hydrolic detention time, including RAS, is 4.2 hours. We need no additional organic to maintain anaerobic conditions in anaerobic tanks. The downside is that phosphorous is released during anaerobic digestion at another facility.

Phosphorus removal in most cases is required today; usually a concentration of CP,e<1–2 mg L–1 has to be maintained. Excess phosphorous is removed by simultaneous precipitation, frequently in combination with enhanced biological phosphorous removal. With a mixing tank upstream of the biological reactor for nitrification–denitrification with a retention time of 15 to a maximum of 30 min, an enhanced biological phosphorous uptake of XP,enh ~0.005–0.007·CCOD,0 can be achieved. For the buildup of heterotrophic biomass, XP,BM = 0.005·CCOD,0 can be assumed. Under these conditions, the concentration of phosphorous to be precipitatedis calculated with Eq. 29:XP,prec = CP,0 – CP,e – XP,BM – XP,enh (29)The effluent concentration should be assumed to be CP,e = 0.6–0.7 of the effluent standard to be maintained. The average requirement for precipitant is calculat-3.2 Technological and microbiological aspects 95 Fig. 3.11 Nitrate to be denitrified as a function of the influent COD and the type of denitrification process. ed as 1.5 mol Me3+ mol–1 XP,prec. With these requirements, Eqs. 30 and 31 are obtained:Precipitation with iron: 2.7 mg SFe mg–1 XP,prec,Fe (30)Precipitation with aluminum: 1.3 mg SAl mg–1 XP,prec,Al (31)The additional mass of excess sludge resulting from phosphorous removal can be calculated with Eq. 32:Mexc,P = Qd · (3 · XP,enh + 6.8 · XP,prec,Fe + 5.3 · XP,prec,Al)/1000 (32)